If a_1,a_2,a_3, and b_1,b_2,b_3, are A.P.s, and a_1=2, a_10=3, a_1b_1=1=a_10b_10, then a_4…

Question

If $a_1,a_2,a_3,\ldots$ and $b_1,b_2,b_3,\ldots$ are A.P.s, and $a_1=2$, $a_{10}=3$, $a_1b_1=1=a_{10}b_{10}$, then $a_4b_4$ is equal to:

Accepted Answer

$a_{10}=2+9d_1=3 \Rightarrow d_1=\dfrac19$; so $a_4 = 2+3\left(\dfrac19\right)=\dfrac73$.

$b_1=\dfrac{1}{a_1}=\dfrac12$, $b_{10}=\dfrac{1}{a_{10}}=\dfrac13$; $b_{10}=b_1+9d_2 \Rightarrow d_2 = -\dfrac{1}{54}$; so $b_4 = \dfrac12+3\left(-\dfrac{1}{54}\right)=\dfrac49$.

$$a_4b_4 = \frac73\times\frac49 = \frac{28}{27}.$$

Solutions · 1